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~christina~
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Homework Statement
x(t)= (7.00m/s)t + (7.00m/s^2)t^2 - (2.00m/s^3)t^3
a) find the expression for the velocity as a function of time.
b) plot the graphs of the position, velocity and acceleration as accleration vs time for the interval given.
c) At what time(s) btw t=0 and t= 3.00 is the particle at rest?
d) for each time calculated in part c) is the acceleration of the particle positive or negative?
e) calculate the accelration of the particle when the velcocity of the particle is at maximum.
f) does the particle have uniform acclerated motion? explain.
Homework Equations
x(t)= (7.00m/s)t + (7.00m/s^2)t^2 - (2.00m/s^3)t^3
The Attempt at a Solution
a) I need to see if my thinking is correct as well as what I did...
v(t)= x'(t)= 7.00m + (14.00m/s)t - (6.00m/s^2)t^2
rearranged:v(t)= -(6.00m/s^2)t^2 + (14.00m/s)t + 7.00m
b) for plotting the graph of velocity and acceleration vs time it would be the equation for my piece and there would be 2 graphs right?
c.) yes I found it to be when the graph v is flat at the top of the parabola..I see that when plug into my calculator and that is when t or x = 1.027 to t or x= 1.311 where v(t)= 15.11 (the flat part = when the accleration = to 0)
d.) If the velocity is 0 wouldn't the acceleration be constant and thus would be positive ?
For this one I'm not sure ...
e.) How would I find the maximum velocity? would that be the max point in the graph? I assume that is also when the acceleration of the particle is 0 since that is the highest point in the v vs t graph but also where the acceleration is 0.
f.) I'm confused about this...wouldnt' the acceleration be 0 when the particle's velocity is max??
g.) I say that the particle does not have uniform accelerated motion since it first starts out as a slanted line toward the right and it is accelerating at a constant rate then it levels off and eventually slants down so the acceleration goes down.
Is this alright??
I'm not sure about d, e, and f
Thank you =D